P is a point on the line joining the points $(3,5,-1)$ and $(6,3,-2)$. If $y$ coordinate of point P is 2 , then $x$ coordinate will be

Let the points be $A(3, 5, -1)$ and $B(6, 3, -2)$. The line passing through $A$ and $B$ can be represented in parametric form. Let $P$ divide the segment $AB$ in the ratio $k:1$.

The coordinates of $P$ are given by the section formula:

$P = \left( \dfrac{6k + 3}{k + 1}, \dfrac{3k + 5}{k + 1}, \dfrac{-2k - 1}{k + 1} \right)$

Given that the $y$-coordinate of $P$ is $2$, we set the $y$-component equal to $2$:

$\dfrac{3k + 5}{k + 1} = 2$

$3k + 5 = 2(k + 1)$

$3k + 5 = 2k + 2$

$k = -3$

Now, substitute $k = -3$ into the expression for the $x$-coordinate of $P$:

$x = \dfrac{6(-3) + 3}{-3 + 1}$

$x = \dfrac{-18 + 3}{-2}$

$x = \dfrac{-15}{-2} = \dfrac{15}{2}$

Answer: $\dfrac{15}{2}$